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خلاصه

توضیحVariations of the Fourier transform.tif	English: Illustration of using Dirac comb functions and the convolution theorem to model the effects of sampling and/or periodic summation. At lower left is a DTFT, the spectral result of sampling s(t) at intervals of T. The spectral sequences at (a) upper right and (b) lower right are respectively computed from (a) one cycle of the periodic summation of s(t) and (b) one cycle of the periodic summation of the s(nT) sequence. The respective formulas are (a) the Fourier series integral and (b) the DFT summation. The relative computational ease of the DFT sequence and the insight it gives into S(f) make it a popular analysis tool.
تاریخ	‏۱۳ دسامبر ۲۰۱۱‏
منبع	اثر شخصی
پدیدآور	Bob K
اجازه‌نامه (استفادهٔ مجدد از این پرونده)	من، صاحب حقوق قانونی این اثر، به این وسیله این اثر را تحث اجازه‌نامهٔ ذیل منتشر می‌کنم:   این پرونده تحت CC0 1.0 Universal Public Domain Dedication کریتیو کامنز قابل دسترسی است. کسی که اثری را با این سند همراه کرده است، با چشم‌پوشی از تمام حقوق خود نسبت به اثر در جهان تحت قانون کپی‌رایت و همهٔ حقوق قانونی مرتبط یا همسایه‌ای که او در اثر داشته است، تا حد مجاز در قانون، آن را به مالکیت عمومی اهدا کرده است. شما می‌توانید بدون گرفتن اجازه این اثر را تکثیر کنید، تغییر دهید، منتشر کنید یا دوباره ایجاد کنید، حتی اگر مقاصد تجاری داشته باشید. http://creativecommons.org/publicdomain/zero/1.0/deed.enCC0Creative Commons Zero, Public Domain Dedicationfalsefalse
دیگر نسخه‌ها	File:Fourier_transform,_Fourier_series,_DTFT,_DFT.svg, File:Fourier_transform,_Fourier_series,_DTFT,_DFT.gif
Source code InfoField	این TIF تصویر برداری با LibreOffice و به‌دست Bob K ساخته شده است
کد منبع InfoField	LibreOffice code Source code pkg load signal graphics_toolkit gnuplot %======================================================= % Consider the Gaussian function e^{-B (nT)^2}, where B is proportional to bandwidth. T = 1; % Choose a relatively small bandwidth, so that one cycle of the DTFT approximates a true Fourier transform. B = 0.1; N = 1024; t = T*(-N/2 : N/2-1); % 1xN y = exp(-B*t.^2); % 1xN % The DTFT has a periodicity of 1/T=1. Sample it at intervals of 1/8N, and compute one full cycle. % Y = fftshift(abs(fft([y zeros(1,7*N)]))); % Or do it this way, for comparison with the sequel: X = [-4*N:4*N-1]; % 1x8N xlimits = [min(X) max(X)]; f = X/(8*N); W = exp(-j*2*pi * t' * f); % Nx1 × 1x8N = Nx8N Y = abs(y * W); % 1xN × Nx8N = 1x8N % Y(1) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π ×-4096/8N × t(n)) } % Y(2) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π ×-4095/8N × t(n)) } % Y(8N) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π × 4095/8N × t(n)) } Y = Y/max(Y); % Resample the function to reduce the DTFT periodicity from 1 to 3/8. T = 8/3; t = T*(-N/2 : N/2-1); % 1xN z = exp(-B*t.^2); % 1xN % Resample the DTFT. W = exp(-j*2*pi * t' * f); % Nx1 × 1x8N = Nx8N Z = abs(z * W); % 1xN × Nx8N = 1x8N Z = Z/max(Z); %======================================================= hfig = figure("position", [1 1 1200 900]); x1 = .08; % left margin for annotation x2 = .02; % right margin dx = .05; % whitespace between plots y1 = .08; % bottom margin y2 = .08; % top margin dy = .12; % vertical space between rows height = (1-y1-y2-dy)/2; % space allocated for each of 2 rows width = (1-x1-dx-x2)/2; % space allocated for each of 2 columns x_origin1 = x1; y_origin1 = 1 -y2 -height; % position of top row y_origin2 = y_origin1 -dy -height; x_origin2 = x_origin1 +dx +width; %======================================================= % Plot the Fourier transform, S(f) subplot("position",[x_origin1 y_origin1 width height]) area(X, Y, "FaceColor", [0 .4 .6]) xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") %xlabel("frequency") %======================================================= % Plot the DTFT subplot("position",[x_origin1 y_origin2 width height]) area(X, Z, "FaceColor", [0 .4 .6]) xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") xlabel("frequency") %======================================================= % Sample S(f) to portray Fourier series coefficients subplot("position",[x_origin2 y_origin1 width height]) stem(X(1:128:end), Y(1:128:end), "-", "Color",[0 .4 .6]); set(findobj("Type","line"),"Marker","none") xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") %xlabel("frequency") box on %======================================================= % Sample the DTFT to portray a DFT FFT_indices = [32:55]*128+1; DFT_indices = [0:31 56:63]*128+1; subplot("position",[x_origin2 y_origin2 width height]) stem(X(DFT_indices), Z(DFT_indices), "-", "Color",[0 .4 .6]); hold on stem(X(FFT_indices), Z(FFT_indices), "-", "Color","red"); set(findobj("Type","line"),"Marker","none") xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") xlabel("frequency") box on %======================================================= % Output (or use the export function on the GNUPlot figure toolbar). print(hfig,"-dtif", "-S1200,900","-color", 'C:\Users\BobK\Fourier transform, Fourier series, DTFT, DFT.tif')

LaTex

${\displaystyle s(t)\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad S(f)\,}$

${\displaystyle S[k]\quad S_{1/T}(f)\quad S_{N}[k]\,}$

${\displaystyle \underbrace {s(t)*\sum _{k=-\infty }^{\infty }\delta (t-kP)} _{\color {BrickRed}\operatorname {rept} _{P}(s(t))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\color {OliveGreen}{\frac {1}{P}}}\underbrace {S(f)\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)} _{\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))}\,}$

${\displaystyle \underbrace {s(t)\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)} _{\operatorname {comb} _{T}(s(t))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{T}}\underbrace {S(f)*\sum _{n=-\infty }^{\infty }\delta (f-n/T)} _{\operatorname {rept} _{1/T}(S(f))}\,}$

${\displaystyle s(t)\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)=\color {Red}\operatorname {comb} _{T}(s(t))\,}$

${\displaystyle {\frac {1}{T}}S(f)*\sum _{n=-\infty }^{\infty }\delta (f-n/T)=\color {Blue}{\frac {1}{T}}\operatorname {rept} _{1/T}(S(f))\,}$

${\displaystyle \underbrace {{\color {BrickRed}\operatorname {rept} _{P}(s(t))}\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)} _{\operatorname {comb} _{T}({\color {BrickRed}\operatorname {rept} _{P}(s(t))})}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{T}}{\color {OliveGreen}{\frac {1}{P}}}\underbrace {{\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))}*\sum _{n=-\infty }^{\infty }\delta (f-n/T)} _{\operatorname {rept} _{1/T}({\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))})}}$

${\displaystyle \underbrace {\operatorname {comb} _{T}(s(t))*\sum _{k=-\infty }^{\infty }\delta (t-kP)} _{\operatorname {rept} _{P}(\operatorname {comb} _{T}(s(t)))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{P}}{\frac {1}{T}}\underbrace {\operatorname {rept} _{1/T}(S(f))\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)} _{\operatorname {comb} _{1/P}(\operatorname {rept} _{1/T}(S(f)))}}$

${\displaystyle {\color {Red}\operatorname {comb} _{T}(s(t))}*\sum _{k=-\infty }^{\infty }\delta (t-kP)=\operatorname {rept} _{P}({\color {Red}\operatorname {comb} _{T}(s(t))})}$

${\displaystyle {\frac {1}{P}}{\color {Blue}{\frac {1}{T}}\operatorname {rept} _{1/T}(S(f))}\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)={\frac {1}{P}}{\color {Blue}{\frac {1}{T}}}\operatorname {comb} _{1/P}({\color {Blue}\operatorname {rept} _{1/T}(S(f))})}$